Let x+y=k be a normal to the parabola  $y^2=12x.$ If P is length of the perpendicular from the focus of the parabola on to this normal then  $4k-2P^2=$

 $y^2=12x$  normal  $x+y=k\to \left(1\right)$

$\Rightarrow a=3$

Normal at $(a{t}^2,2at)$

Equation of normal 

Equation of normal 

$2at(x-a{t}^2)+2a(y-2at)=0\to \left(2\right)$

From (1) and (2) $t=1$

$k=3t^3+6t$

$k=9$

$s(a,0)=(3,0)$

Perpendicular distance p=d $=3\sqrt{2}$

$=4k-2p^2=4\left(9\right)-2{\left(3\sqrt{2}\right)}^2$

$=36-36=0$